Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 5 - Section 5.1 - Exponents - Exercise Set - Page 345: 104



Work Step by Step

Based on the power of a product rule, we know that $(ab)^{n}=a^{n}b^{n}$ (where $n$ is a positive integer and $a$ and $b$ are real numbers). Therefore, $(\frac{3a^{4}}{9b^{5}})^{2}=\frac{3^{2}(a^{4})^{2}}{9^{2}(b^{5})^{2}}=\frac{9(a^{4})^{2}}{81(b^{5})^{2}}=\frac{(a^{4})^{2}}{9(b^{5})^{2}}$. Based on the power rule for exponents, we know that $(a^{m})^{n}=a^{mn}$ (where $m$ and $n$ are positive integers and $a$ is a real number). Therefore, $\frac{(a^{4})^{2}}{9(b^{5})^{2}}=\frac{a^{4\times2}}{9\times b^{5\times2}}=\frac{a^{8}}{9b^{10}}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.